Transcript Lecture 8
Lecture 8
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Lecture 8 – Tuesday 2/5/2013
Block 1:
Block 2:
Block 3:
Block 4:
Mole Balances
Rate Laws
Stoichiometry
Combine
Pressure Drop
Liquid Phase Reactions
Gas Phase Reactions
Engineering Analysis of Pressure Drop
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Pressure Drop in PBRs
Concentration Flow System:
Gas Phase Flow System:
CA
3
FA
CA
FA
0 1 X
FA0 1 X
C 1 X T0 P
A0
T P0
1 X T P0
0 1 X
T0 P
b
b
FA0 B X C A0 B X
FB
a
a T0 P
CB
P
T
1 X T P0
1 X
0
0
T0 P
T P0
T0 P
Pressure Drop in PBRs
Note: Pressure Drop does NOT affect liquid phase reactions
Sample Question:
Analyze the following second order gas phase reaction
that occurs isothermally in a PBR:
AB
Mole Balances
Must use the differential form of the mole balance to
separate variables:
dX
FA0
rA
dW
Rate Laws
Second order in A and irreversible: rA kC A2
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Pressure Drop in PBRs
1 X P T0
CA
CA 0
1 X P0 T
FA
Stoichiometry
Isothermal, T=T0
Combine:
1 X P
CA CA 0
1 X P0
dX kC 1 X P
2
dW
FA0 1 X P0
2
A0
2
2
Need to find (P/P0) as a function of W (or V if you have a PFR)
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Pressure Drop in PBRs
Ergun Equation:
dP
G 1 1501
3
1
.75
G
dz g c D p
Dp
TURBULENT
LAMINAR
Constant mass flow:
m m 0
0 0
0
0
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FT P0 T
0
FT 0 P T0
P0 T
0 (1 X )
P T0
Pressure Drop in PBRs
Variable Density
dP
G
dz 0 g c D p
Let
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P T0 FT 0
0
P0 T FT
P0 T FT
1 1501
3
1.75G
Dp
P T0 FT 0
G
0
0 gc Dp
1 1501
3
1.75G
Dp
Pressure Drop in PBRs
Catalyst Weight
Where
Let
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W zAc b zAc 1 c
b bulk density
c solid catalyst density
porosity (a.k.a., void fraction )
(1 ) solid fraction
0
P0 T FT
dP
dW Ac 1 c P T0 FT 0
2 0
1
Ac 1 c P0
Pressure Drop in PBRs
dy
T FT
dW
2 y T0 FT 0
P
y
P0
We will use this form for single reactions:
d P P0
1 T
1 X
dW
2 P P0 T0
dy
T
1 X
dW
2 y T0
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dy
1 X
dW
2y
Isothermal case
Pressure Drop in PBRs
kC 1 X 2
dX
y
2
dW
FA0 1 X
2
A0
2
dX
dP
f X , P and
f X , P or
dW
dW
dy
f y, X
dW
The two expressions are coupled ordinary differential
equations. We can only solve them simultaneously using
an ODE solver such as Polymath. For the special case of
isothermal operation and epsilon = 0, we can obtain an
analytical solution.
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Polymath will combine the Mole Balances, Rate Laws
and Stoichiometry.
Packed Bed Reactors
For 0
dy
(1 X )
dW 2 y
When W 0 y 1
dy dW
2
y (1 W )
2
y (1 W )1/ 2
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1 Pressure Drop in a PBR
P
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W
2 Concentration Profile in a PBR
CA
P
CA CA 0 1 X
P0
No P
P
W
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3 Reaction Rate in a PBR
-rA
P
rA kC k (1 X )
P0
2
A
2
No P
W
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P
2
4 Conversion in a PBR
X
No P
P
W
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5 Flow Rate in a PBR
P
For 0 :
f
0
P0
0
P
1
No P
W
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P0 T
0 1 X
P T0
T T0
P0
y
P
0
1
f
(1 X ) y
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Example 1:
Gas Phase Reaction in PBR for =0
Gas Phase reaction in PBR with δ = 0 (Analytical
Solution)
A + B 2C
Repeat the previous one with equimolar feed of A
and B and:
kA = 1.5dm6/mol/kg/min CA0 CB0
α = 0.0099 kg-1
Find X at 100 kg
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C A0
CB0
X ?
Example 1:
Gas Phase Reaction in PBR for =0
1) Mole Balance
dX r ' A
dW
FA0
2) Rate Law
r ' A kCACB
3) Stoichiometry
CA CA0 1 X y
CB CA0 1 X y
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Example 1:
Gas Phase Reaction in PBR for =0
dy
dW
2y
W 0
2, ydy dW
y 2 1 W
, y 1
y 1 W
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4) Combine
rA kC 1 X y kC 1 X 1 W
2
A0
2
2
2
A0
dX kCA2 0 1 X 1 W
dW
FA0
2
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2
Example 1:
Gas Phase Reaction in PBR for =0
kCA2 0
dX
1 W dW
2
1 X FA0
kCA2 0
X
W 2
W
1 X
FA0
2
W 0, X 0, W W , X X
X 0.6 with pressure drop
X 0.75 without pressure drop, i.e. 0
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Example 2:
Gas Phase Reaction in PBR for ≠0
The reaction
A + 2B C
is carried out in a packed bed reactor in which there is
pressure drop. The feed is stoichiometric in A and B.
Plot the conversion and pressure ratio y = P/P0 as a
function of catalyst weight up to 100 kg.
Additional Information
kA = 6 dm9/mol2/kg/min
α = 0.02 kg-1
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Example 2:
Gas Phase Reaction in PBR for ≠0
A + 2B C
1) Mole Balance
dX rA
dW FA0
2) Rate Law
rA kCACB2
3) Stoichiometry: Gas, Isothermal
P0
0 1 X
P
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1 X
C A C A0
y
1 X
Example 2:
Gas Phase Reaction in PBR for ≠0
B 2 X
4) CB C A0 1 X y
dy
1 X
5)
dW
2y
1 X
6) f
0
y
1
2
7) y A0 [1 1 2] [2]
3
3
C A0 2, FA0 2, k 6, 0.02
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Initial values: W=0, X=0, y=1
Final values: W=100
Combine with Polymath.
If δ≠0, polymath must be used to solve.
Example 2:
Gas Phase Reaction in PBR for ≠0
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Example 2:
Gas Phase Reaction in PBR for ≠0
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T = T0
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Robert the Worrier wonders: What if we increase
the catalyst size by a factor of 2?
Pressure Drop
Engineering Analysis
𝜌0 = 𝑀𝑊 ∗ 𝐶𝑇0 =
𝑀𝑊 ∗ 𝑃0
𝑅𝑇0
2𝑅𝑇0
150 1 − 𝜙 𝜇
𝛼=
𝐺
+ 1.75𝐺
𝐷𝑃
𝐴𝐶 𝜌𝐶 𝑔𝐶 𝑃02 𝐷𝑃 𝜙 3 𝑀𝑊
1
𝛼≈
𝑃0
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2
Pressure Drop
Engineering Analysis
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Pressure Drop
Engineering Analysis
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Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
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End of Lecture 8
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Pressure Drop - Summary
Pressure Drop
Liquid Phase Reactions
Pressure Drop does not affect concentrations in lquid phase
reactions.
Gas Phase Reactions
Epsilon does not equal to zero
d(P)/d(W)=…
Polymath will combine with d(X)/d(W) =… for you
Epsilon = 0 and isothermal
P=f(W)
Combine then separate variables (X,W) and integrate
Engineering Analysis of Pressure Drop
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Pressure Change – Molar Flow Rate
FT P0 T
0
FT 0 P T0
dP
dW
A c 1 c
FT T
0
FT 0 T0
dy
dW
yP0 A c 1 c
dy
FT T
dW
2 y FT 0 T0
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P0 A C 1 C
Use for heat effects,
multiple rxns
FT
1 X Isothermal: T = T0
FT 0
dX
1 X
dW
2y
Example 1:
Gas Phase Reaction in PBR for =0
A + B 2C
dm6
1
k 1.5
, 0.0099kg
m ol kg min
Case 1:
W 100kg , X ? , P ?
Case 2: DP 2 DP1
C A0
CB0
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, C B 0 C A0
1
, P02 P01 ,
2
X ? , P?
X ?
P?
PBR
dX
FA0
r ' A
dW
rA kC ACB
FA
CA
y
FT
C A CB
0 and T T0 y (1 W )1/2
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